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Engineering Fundamentals and Problem Solving, 6e

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Engineering Fundamentals and Problem Solving, 6e. Chapter 17 Electrical Circuits. Chapter Objectives. Compute the equivalent resistance of resistors in series and in parallel Apply Ohm’s law to a resistive circuit

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Engineering Fundamentals and Problem Solving, 6e

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  1. Engineering Fundamentals and Problem Solving, 6e Chapter 17 Electrical Circuits

  2. Chapter Objectives Compute the equivalent resistance of resistors in series and in parallel Apply Ohm’s law to a resistive circuit Determine the power provided to a DC circuit and the power used by circuit components Use Kirchhoff’s laws to solve resistive networks Utilize mesh currents to solve resistive networks

  3. Simple DC Electric Circuit and Symbols

  4. Ohm’s Law Potential = Current X Resistance Where V =Potential in volts R =Resistance in ohms I = Current in amperes

  5. Resistors in Series V1 V2 V3 VT

  6. Resistors in Parallel VT

  7. DC Electric Power

  8. Kirchhoff’s Laws Kirchoff’s voltage law • “The algebraic sum of all the voltages (potential drops) around any closed loop in a network equals zero.”  Vdrops= 0 Kirchoff’s current law • “The algebraic sum of all of the currents coming into a node (junction) in a network must be zero.”  Inode= 0

  9. Circuit Example 17.7 Given the following circuit, determine the currents Ix, Iy, and Iz.

  10. Circuit Example cont’d From Kirchhoff’s current law at point A Iy= Ix+ Iz From Kirchhoff’s voltage law around left loop - Iy(2) + 14 – Ix(4) = 0 Around right loop - Iy(2) + 12 – Iz(6) = 0 Results in: Ix = 2A, Iy = 3A, Iz = 1A

  11. Mesh Currents • A node is a specific point or location within a circuit where two or more components are connected. • A branch is a path that connects two nodes. • A mesh is a loop that does not contain any other loops within itself. • Mesh currents • Exist only in the perimeter of the mesh • Selected clockwise for each mesh • Travel all the way around the mesh

  12. Mesh Current Example V1 V2 Write the mesh current equations for this circuit. V1 – IaR1 – (Ia – Ib)R3 = 0 -V2 – (Ib – Ia)R3 – IaR2 = 0

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